Question

If
$w \alpha = ( \alpha |x \: e \: w) \: then \: w3 \: n \: w5 =$
option 1) W3 2) W5
3) W8 4) W15
Don't write only answer give explanation I know correct answer is 4) W15 But How ?? ​

Answer 1

Correct Question:-

If $\sf{W_{\alpha}=\{x:\alpha\mid x,\ x\in\mathbb{W}\},}$ then $\sf{W_3\cap W_5=}$

$\sf{1.\quad W_3}$

$\sf{2.\quad W_5}$

$\sf{3.\quad W_8}$

$\sf{4.\quad W_{15}}$

Solution:-

By definition,

$\longrightarrow\sf{W_3=\{x:3\mid x,\ x\in\mathbb{W}\}}$

And,

$\longrightarrow\sf{W_5=\{x:5\mid x,\ x\in\mathbb{W}\}}$

For two non - empty sets $\sf{A}$ and $\sf{B}$ we know that,

$\longrightarrow\sf{A\cap B=\{x:x\in A\ and\ x\in B\}}$

Hence,

$\longrightarrow\sf{W_3\cap W_5=\{x:3\mid x\ and\ 5\mid x,\ x\in\mathbb{W}\}\quad\quad\dots(1)}$

By divisibility rule, if a number is exactly divisible by 3 as well as 5, then it is exactly divisible by 15 too.

$\longrightarrow\sf{3\mid x\ and\ 5\mid x\implies15\mid x}$

Therefore (1) becomes,

$\longrightarrow\sf{W_3\cap W_5=\{x:15\mid x,\ x\in\mathbb{W}\}}$

By definition,

$\longrightarrow\underline{\underline{\sf{W_3\cap W_5=W_{15}}}}$

Hence 4th option is the answer.